ICSE Board

Class IX Mathematics

Paper 1 – Solution

SECTION – A (40 Marks)

Q. 1.

(a) Here,

n

2

C.I. Rs. 287, r 5% p.a., n 2 years

rC.I. P 1 1 287

100

5P 1 1 287

100

2

2

105P 1 = 287

100

21P 1 = 287

20

441P 1 = 287

400

441 400P = 287

400

41P 287

400

287 400P 2800

41

Thus P = Rs. 2,800

(b) Let us assume that 2 is a rational number.

Then, 2 p

q ….(1)

Where p and q are integers, co-prime to each other and q 0.

On squaring both sides, we get2

2 2

2

p2 p 2q ....(2)

q

By equation (2), we can say that p2 is an even integer.

p is also an even integer ( the square of an even integer is always even)

Let p = 2k, where k is an integer.

1

From (2), 2 2

2 2

2 2

2 2

p = 2q

(2k) = 2q

4k = 2q

q = 2k

q2 is an even integer, q is also an even integer.

Thus, p and q have a common factor 2 which contradicts the hypothesis that p, q are

co-prime to each other.

2 is an irrational number.

(c) o o

o o o o o

o o o

o o o o o o o

oo o

o o o o o o

cos37 .cosec53

tan5 .tan25 .tan45 .tan65 .tan85

cos37 .cosec(90 37 )

tan5 .tan25 .tan45 .tan(90 25 ).tan(90 5 )

cosec(90 ) seccos37 .sec37

tan5 .tan25 .tan45 .cot 25 .cot5 tan(90 ) cot

o o o oo

1sec1

costan5 .cot5 .tan25 .cot 25 .1

tan45 1

1 1cot

1 tan

1

Q. 2

(a) In ABD and BCD

mA = mC = 90

AB = DC [Given]

BD = BD [Common]

ABD CDB [R.H.S.]

CBD = ADB [C.P.C.T.]

CBD = x = 50

Also, BC = AD [C.P.C.T.]

3y – 20 = y – 10

3y – y = 20 – 10

2y = 10

y = 5 cm

2

(b) 2 log 1

32

log 16 + log 12 1

2 2= log3 log(16) + log12

= log9 log4 + log12

9 × 12= log

4

= log27

(c) Steps of construction for constructing parallelogram:

1) Draw a line AB of measure 6 cm.

2) Draw an angle of measure 45° at point A such that DAB = 45° and AD = 5 cm.

3) Now draw a line CD parallel to line AB of measure 6 cm.

4) Join BC. Join diagonals AC and BD. Let them intersect at O.

Thus, ABCD is the required parallelogram.

Q. 3.

(a)

222 3 233 0

2 2

22

8 1 2 17 1 Since a = 1

27 3 3 3

2 1= 1

3 3

3= 3 1

2

9= 9 1

4

9= 10

4

9 40=

4

31=

4

3

(b) Given, (2a + b, a – 2b) = (7, 6)

2a b 7 ....(1)

a 2b 6 ....(2)

Multiplying equation (1) by 2, we get

4a 2b 14 ....(3)

Adding equations (2) and (3), we get

5a 20

a 4

Substituting a 4 in equation (1), we get

2(4) b 7

b 1

a 4 and b 1

(c) Let A (0, 0), B (5, 0), C (8, 4) and D (3, 4)

2 2

2 2

2 2

2 2

2 2

2 2

AB = (5 0) + (0 0) = 25 + 0 = 5

BC = (8 5) + (4 0) = 9 + 16 = 25 = 5

CD = (8 3) + (4 4) = 25 + 0 = 5

DA = (0 3) + (0 4) = 9 + 16 = 25 = 5

AC = (8 0) + (4 0) = 64 + 16 = 80 = 4 5

BD = (3 5) + (4 0) = 4

+ 16 = 20 = 2 5

Now, AB = BC = CD = DA and AC BD.

Hence, ABCD is a rhombus.

Q. 4.

(a) Given side of an equilateral triangle = a

Let, AD DC ∴ BD = DC = a

2

In right ABD, 2 2 2AB AD BD [Using Pythagoras theorem]

2 2 22 2 2 2

2

a a 3a a 3a = AD + AD = a = AD =

2 4 4 2

1 1 1 a 3 3Now, Area of ΔABC = × Base × height = × BC × AD = × a × = a

2 2 2 2 4

1 1Area of rhombus = × AC × BD = × 4 5 × 2 5 20 sq. units

2 2

4

(b) We know, each exterior angle =o360

n

o o

o

o

2

o2

2

2

2

360 3606

n n 2

n 2 n360 6

n n 2

360 26

n 2n

360 2n 2n

6

n 2n 120

n 2n 120 0

n 12n 10n 120 0

n n 12 10 n 12 0

n 12 n 10 0

n 12 or n 10

Neglecting n = −12 as number of sides cannot be negative.

n = 10

Thus, number of sides are 10.

2 o

2 o 2 o

4 1(c) + tan 45

tan 60 cos 30

2

2 2

4 1= + 1

3 3/2

4 4

= + 13 3

5=

3

5

SECTION – B (40 Marks)

Q. 5

(a) 3x – 5y + 1 = 0

3x 1

y5

x 1 3 –2

y 0.8 2 –1

And 2x – y + 3 =0

y = 2x + 3

x 0 1 -1

y 3 5 1

From the graph, we find that the two lines intersect at the point (2, 1)

Therefore, the solution is x = 2, y = 1

(b) Let his starting salary be Rs. x and the fixed annual increment be Rs. y

Salary after 4 years = x + 4y

According to question,

x + 4y = 1500 ….(i)

And salary after 10 years = x + 10y

x + 10y = 1800 ….(ii)

Subtracting (i) from (ii), we get

6y = 300

y = 50

Substituting the value of y in (i), we get

x + 4y = 1500

x + 4 50 = 1500

x = 1300

Starting salary = Rs. 1300 and Annual increment = Rs. 50

6

(c)

2

2

1 1 2 1 2 1x 2 1

2 12 1 2 1 2 1

x 2 1 2 1 2 2 3 2 2

Therefore,

22

22

1 1x 6 2 1 6

x 2 1

13 2 2 6

3 2 2

3 2 2 6 3 2 2

0

Q. 6.

(a) Let the sum be Rs. P, A = Rs. 3630, r = 10%, n = 2 years

We know that,n

2

2

rA P 1

100

103630 P 1

100

1103630 P

100

3630 10 10P

11 11

P Rs. 3000

(b) In right PQS,

2 2 2PS QS PQ [By Pythagoras theorem]

2 22PS 10 6 100 36 64

PS 8 cm

RS = RQ + QS = 9 + 16 = 15 cm

2 2 2

2 22

Now in PSR,

PR PS RS

PR 8 15 64 225 289

PR 17cm

7

(c) Consider the following figure:

2 2 2

2 2 2

2

2

2 2

Distance of smaller chord AB from centre of circle 4 cm

i.e., OM 4 cm

AB 6MB 3 cm

2 2

In OMB,

OM MB OB

4 3 OB

16 9 OB

OB 25

OB 25 5 cm

In OND,

OD OB 5 cm ....(radii of the same circle)

CD 8ND 4 cm

2 2

ON ND

2

2 2 2

2 2 2

OD

ON OD ND

ON 5 4 25 16 9

ON 9 3 cm

Q. 7.

(a) Arranging the numbers in ascending order:

1, 2, 3, 3, 3, 4, 5, 5, 6, 7

Number of terms = n = 10 (even)

Mean =1 + 2 + 3 + 3 + 3 + 4 + 5 + 5 + 6 + 7

10

393.9

10

And median = th th5 term + 6 term

2

3 4 73.5

2 2

8

(b) Area of room = (8 5) m2 = 40 m2

Let the length of carpet be ‘x’ m.

Area of carpet = l b

280x 0.80x m

100

Area of carpet = Area of floor

0.80x = 40

40x = × 100 = 50 m

80

Cost of carpet = 50 Rs. 22.50 = Rs. 1125

(c) PR = PQ [Given]

1 = 2 [angles opposite to equal sides]

Exterior angle of a triangle is greater than any one of the

interior opposite angles.

1 > 3

2 > 3 [ 1 = 2]

In PSR, R > S

PS > PR [sides opposite to greater angle is longer]

PS > PQ [ PR = PQ]

Q. 8.

(a) Length (l) of the greenhouse = 30 cm

Breadth (b) of the greenhouse = 25 cm

Height (h) of the greenhouse = 25 cm

i. Total surface area of the greenhouse = 2[lb + lh + bh]

= [2(30 × 25 + 30 × 25 + 25 × 25)]

= [2(750 + 750 + 625)]

= (2 × 2125)

= 4250 cm2

Thus, the area of the glass is 4250 cm2.

ii. Total length of tape = 4(l + b + h)

= [4(30 + 25 + 25)] cm

= 320 cm

Thus, 320 cm of tape is required for all the 12 edges.

9

(b) Given,

o

1. AOC is the diameter

12. Arc AXB = Arc BYC

2

1From Arc AXB = Arc BYC we can see that

2

Arc AXB : Arc BYC = 1: 2

BOA : BOC = 1 : 2

Since AOC is the diameter of the circle,

hence, AOC = 180

Now,

Assume tha

o o

o

o o

t BOA = x and BOC = 2x

AOC = BOA + BOC = 180

x + 2x = 180

3x 180

x = 60

Hence BOA = 60 and BOC = 120

(c)

Age in years 10-20 20-30 30-40 40-50 50-60 60-70

Number of patients 90 40 60 20 120 30

Take class intervals i.e. age in years along x-axis and number of patients of width equal to the size of the class intervals and height equal to the corresponding frequencies to get the required histogram.

In order to draw frequency polygon, we take imaginary intervals 0-10 at the beginning and 70-80 at the end each with frequency zero and join the mid-points of top of the rectangles. Thus, we obtain a complete frequency polygon, shown below:

10

Q. 9.

(a)2

2

2

2

2

2

2cos sin 2 0

2cos sin 2

sin 2 2cos

sin 2(1 cos )

sin 2 sin

sin 1

sin 2

1sin

2

sin sin30

30

(b) 11 1yx zAssume p p p k

11 1yx z

x y z

x y z

x y z 0

p k,p k,p k

p k ,q k ,r k

Also, pqr 1

k k k 1

k k

x y z 0

(c) From the given figure,

1XD AD

2 ( X is the midpoint of AD)

And 1

BY BC2

( Y is the midpoint of BC)

XD = BY (as AD = BC opposite sides of ∥gm)

Also, XD ∥ BY ( AD ∥ BC opp. Sides of ∥gm)

XBYD is a parallelogram (Opposite sides are equal and parallel)

In AFD, X is the midpoint of AD and XE ∥ DF

E is the midpoint of AF

AE = EF ---- (i)

Now in CEB, Y is the midpoint of BC and YF ∥ BE.

F is the midpoint of CE

EF = FC ---- (ii)

From (i) and (ii)

AE = EF = FE [Hence proved] 11

Q. 10.

(a)

On comparing this with P 35 Q , 1

P2

and 1

Q2

(c) 3 cos A – 4 sin A = 0

3cosA = 4sinA

sinA 3=

cosA 4

3tanA =

4

2 2 2

By Pythagoras theorem,

AC = AB + BC2 2= 4 + 3

= 16 + 9 = 25

AC = 25 = 5

BC 3sinA = =

AC 5

AB 4

cos AAC 5

sin A 2cosA

3cosA sin A=

3 4 3 8 112

95 5 5 5 54 3 12 3 11 535 5 5 5 9

12

(c) i. 1

a + = 4,a

On squaring, we get

2

2

1a

a + 2 = 16

2

2

1a

a = 16 2 = 14

ii. By part (i), we have 2

2

1a

a = 14

On squaring, we get

4

4

1a

a + 2 = 196

4

4

1a

a = 196 2 = 194

Q. 11.

(a) Given: In ∆ABC, AD is the median

To prove: Area of ABD = Area of ADC

Construction: We draw AE BC

Proof:

Area of ABD = 1

BD AE2

1[ Area base height]

2

Similarly, area of ADC = 1

DC AE2

Here, we have BD = DC

Area of ABD = Area of ADC

Hence proved.

(b) Given, Area of PQR = 44.8 cm2

Since QL is the median and Median divides triangle into two

triangles of equal areas

Area of LQR = Area of PQL = 22.4 cm2

In QLR, LM is the median,

Area of LMR = 2122.4 11.2cm

2

(c) x3 – 3x2 – x + 3

= x2 (x – 3) – 1(x – 3)

= (x – 3)(x2 – 1)

= (x – 3)(x – 1)(x + 1)

13